 Hello, and welcome to a screencast about L'Hopital's rule. So L'Hopital's is a rule that says if f and g are differentiable functions, at a certain point, let's call it x equals a. And we know that f of a and f of g are both 0. And also, g prime of a cannot be 0. Then what L'Hopital's rule says is if you take the limit as x approaches a of your functions, that's the same thing as your derivatives evaluated at that point. It's also the same thing as the limit as x approaches a of your derivatives, assuming everything is continuous. So the first function we are going to look at today is x cubed minus 5x squared plus 11x minus 10 all over x squared minus 4. So I went ahead and graphed that in GeoAjibra. And I include the graph here. And today, we're going to take a look at the limit as x approaches 2. OK, so looking at GeoAjibra here, so here's x equals 2. If we were to eyeball our limit, I don't know, it's something less than 1. Somewhere, maybe bigger than a half, but less than 1. Kind of hard to tell. So now we've got to go through and check our conditions here to see if we can use L'Hopital's rule and if it does actually make sense. So if we were to plug 2 into this function, so let's look at the numerator and denominator separately. If we were to plug 2 in, we'd end up with, so this is just kind of checking some things here. So let's see, we'd have 2 cubed minus 5 times 2 squared plus 11 times 2 minus 10. So remember, to build a use L'Hopital's rule, we need to have that f of a, which in this case is f of 2. Our numerator function evaluated at 2 has to give us 0. Otherwise, we can't use this. Let's also check our denominator. So here, if we plug in a 2, so 2 squared minus 4. Okay, well, certainly the denominator's 0, definitely. I'm going to put this in quotes too. Only because we're hopefully going to end up with an indeterminate form, which was what L'Hopital's is good for. All right, anyway, we simplify this numerator. We end up with, multiply all this stuff out. We get 8 minus 20 plus 22 minus 10, and sure enough, that gives us a 0 as well. Okay, so this is definitely an indeterminate form. 0 over 0 doesn't necessarily mean it's undefined. Doesn't mean it's 1, doesn't mean it's 0. We don't know. So that's what we're going to have to use L'Hopital's then. So, and we know that they're good because, differentiable? Sure. Does the numerator's a polynomial? The denominator's a polynomial. We also know they're continuous because they're both polynomials again. So we're certainly good here. Okay, I'm going to write the little, or the abbreviation here L'H, just so we know we've used L'Hopital's. So that says then we can go ahead and evaluate these at the derivatives. So let's take the derivative of the numerator. So that's going to give us 3x squared minus 10x plus 11. And we can take the derivative of the denominator separately. That gives us a grand total of 2x. So we've now used L'Hopital's in one step. Let's go ahead and see if we can now evaluate this new function at 2 and see if we get a value that makes any sense. Or we made up with 0 over 0 again, which means we'll have to use L'Hopital's again. But if we go ahead and plug in our values, we end up getting, let's see, 3 times 2 squared minus 10 times 2 plus 11. This is kind of our little scratch work off to the side. 2 times 2. So we can definitely tell the denominator is not going to give us 0. So let's go ahead and write out what that value is going to be, which is 4. And if we simplify the numerator, I believe that gives us a grand total of 3. You notice my limit went away, because now this is a function that is good at 2. It doesn't have any problems there. And sure enough, our value of something less than 1, but bigger than 1 half was certainly true. So my limit is 3 4's. Great, let's take a look at another function that's not a polynomial. So this one is a cosine function. Excuse me. So if you take the limit as x approaches 0, cosine of x minus 1 all over x squared. So again, if we take a look at a graph on Geoge, or your calculator, whatever floats your boat, this one looks like it might be approaching a half, or negative a half, sorry, because we're down here in the negative area. But who knows if that's exactly it? Who knows if there's something funky going on here that I can't quite see with the way I've got my graph zoomed? You just never know. So again, let's go ahead and take a look at what's happening to our numerator and denominator at 0 to see if we can use L'Hopital's, see if we get something that's indeterminate. So if I evaluate the numerator at 0, so cosine of 0 minus 1, that's going to give me 1 minus 1, which is 0. And if I go ahead and plug a 0 in the denominator, obviously 0 squared is also 0. So again, I'm going to go ahead and write this in quotes as 0 over 0. So this is a case we can use L'Hopital's on. So I'm going to go ahead and do that. I'm going to write an L'H again to indicate that I've done L'Hopital's. Limit x approaches 0. And let's go and do the derivative of our numerator. So the derivative of cosine is negative sine of x. Minus 1, obviously, that piece goes away. The derivative of our denominator, x squared is 2x. OK, not too bad. So these are pretty easy when you do them separately. OK, now let's take a look at what happens to these functions as x approaches 0. Well, I see a 0 in the denominator again. That's never a good sign. OK, so let's go ahead and check what the numerator is to, and let's see if we end up with 0 over 0. And sure enough, sine of 0 is 0. Negative in front doesn't make a darn bit of difference. So we have our 0 over 0 case again. So we can use L'Hopital's one more time. We may need it again, but you just never know. Limit x approaches 0. The derivative of our numerator is negative cosine. So this is where all your derivatives of all your different functions are going to come in handy. Derivative of our denominator is 2. OK, so obviously we're not going to have L'Hopital's again because our denominators are constant. And if we plug 0 into our numerator, we're going to get negative 1. So sure enough, our limit is, as we thought when we eyeballed our graph, negative 1 half. But now we actually have some algebra and some calculus here that will prove the fact that our limit is negative 1 half instead of just eyeballing off the graph. Thank you very much.